What is the sum from n=1 to infinity of 2(0.7)^{n-1} ?

The sum from n=1 to infinity of 2(0.7)^{n-1} is converges

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sum from n=1 to infinity of 2(0.7)^{n-1}

n = 1 \sum \infty 2 (0.7)^{n - 1}

converges

Solution steps

n = 1 \sum \infty 2 \cdot 0. 7^{n - 1}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 2 \cdot n = 1 \sum \infty 0. 7^{n - 1}

Apply Series Ratio Test:

converges

= 2 converges

= converges

n = 0 \sum \infty cos (n) π

n = 1 \sum \infty \frac{1}{3 ^{n} n !}

n = 1 \sum \infty (\frac{2}{n})^{3}

n = 1 \sum \infty cos (\frac{3}{n})

n = 0 \sum \infty (\frac{1}{π})^{n}

What is the sum from n=1 to infinity of 2(0.7)^{n-1} ?
The sum from n=1 to infinity of 2(0.7)^{n-1} is converges